Solution

a) From Ampere's law, we find the magnetic field induction $B$ $$F_1 = Il \cdot B \Rightarrow B = \frac{F}{Il}\tag{1}$$ in this case, the numbers acting on the part of the wire with currents $I_1$ and $I_2$, respectively $$F_1 = I_1l \cdot B; \quad F_2 = I_1L\cdot B$$ We substitute $(1)$: $$F_1= \frac{I_1}{I}F; \quad F_2 = \frac{I_1L}{Il}F$$ The force acting on a piece of wire bent at an angle $\varphi$ can be found from the cosine law $$|\vec{F_p}| = |\vec{F_1} + \vec{F_2}| = \sqrt{{F_1}^2+{F_2}^2+2 F_1 F_2\cdot\cos\varphi}$$ $$\boxed{|\vec{F_p}| = F\frac{I_1}{I}\sqrt{1+\frac{L^2}{l^2} + 2\frac{L}{l}\cdot\cos\varphi}}$$ b) For a conductor made of a semicircle, we will use the same formula for the magnetic field induction $B$ $$B = \frac{F}{Il}$$ On a small part of a conductor of length $dl$, the Ampere force acts $$dF = B I_2~dl = B I_2R~d\theta$$ To find the force $F$, we integrate the forces acting on each part of the semiring. $$F = \int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos\theta dF = BI_2R\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos\theta~d\theta$$ After mathematical transformations, we obtain the force acting on the semiring $$\boxed{F = 2BI_2R = 2F\frac{I_2R}{Il}}$$